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Homework Help: Proportionality of T to R, m1 and m2

  1. Apr 18, 2005 #1
    We have done an experimnet latly in which we did an investigation on

    m1Ug=4Pie^2m(2)r m2/ T^2

    We drew the graphs of the values that we found, then we drew the Log values to fing Gradiant to find the power of proportionality.

    in our experiment gradiant of my Log values were T prop m2^.53 T prop R^.508 and T prop m1^-.35

    using math I got got

    T^2 m1Ug =4pie^2 m2 r
    T^2 = 4pie^2 m2 r / m1Ug
    T = 2pie Root m2r/ root m1 Ug

    Which makes it

    T prop m1^.5 or root m2
    same for R
    and ^-.5 for m1 or 1/root m1

    does that seem good? even though keplers third law stats T^2 is prop to r^3
    Last edited: Apr 18, 2005
  2. jcsd
  3. Apr 18, 2005 #2


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    It's hard to say! You didn't tell us what this equation was supposed to be so I don't know what Kepler's third law has to do with it! I would prefer you used "pi" rather than "pie" for [tex]\pi[/tex]. I thought at first this involved the exponential
    e^(2m)! My real problem is distinguishing between "m(2)" and "m2". Are they intended to be the same thing? And are "r" and "R" supposed to mean the same thing?
  4. Apr 18, 2005 #3
    Sorry for the mistake.

    Here is the equasion

    F=4pi^2m2r/T^2 F is m1Ug.

    so now m1Ug=4pi^2m2r /T^2

    I did some work out and mate T to be

    T proportional to m1^-.5
    T prop to m2^.5
    and T prop to r^.5

    as T= 2pi sqroot m2r / sqroot m1Ug

    is that right?

    here is a link

    http://users.tpg.com.au/timedil/cent.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  5. Apr 18, 2005 #4


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    Science Advisor

    Yes, that is correct.
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